A and B throw a die alternatively till one of them gets a number more than 4 and wins the game. Then the probability of winning the game by B, if A starts first:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) - $\frac{2}{5}$
P(Success) = P(S) = P(getting greater than 4) = $\frac{1}{3}$
P(Failure) = P(F) = $\frac{2}{3}$
P(B success) = A fails B succeeds + A fails B fails A fails B succeeds + .........
= A fails succeeds (1 + B fails A fails + B fails A fails B fails A fails .....)
$=\frac{2}{3}×\frac{1}{3}(1+(\frac{2}{3})^2+(\frac{2}{3})^4+(\frac{2}{3})^6.....∞)$
$=\frac{2}{9}×\frac{1}{1-(\frac{2}{3})^2}=\frac{2}{5}$